Question

Two radioactive substances $A$ and $B$ have decay constants $5\lambda$ and $\lambda$ respectively. At $t = 0$, they have the same number of nuclei. The ratio of number of nuclei of $A$ to those of $B$ will be $(1/e)^{2}$ after a time interval

• A. $4\, \lambda$
• B. $2\, \lambda$
• C. $1/2\,\lambda$
• D. $1/4\,\lambda$

Answer 1

Answer: (C)

$1/2\,\lambda$

Answer 2

Given : $\lambda_{A}=5\lambda, \lambda_{B}=\lambda$ At $t=0, \left(N_{0}\right)_{A}=\left(N_{0}\right)_{B}$ At time $t, \frac{N_{A}}{N_{B}}=\left(\frac{1}{e}\right)^{2}$ According to radioactive decay, $\frac{N}{N_{0}}=e^{-\lambda t}$ $\therefore \frac{N_{A}}{\left(N_{0}\right)_{A}}=e^{-\lambda_{A^t}}$ and $\frac{N_{B}}{\left(N_{0}\right)_{B}}=e^{-\lambda_{B^t}}$ Divide $\left(i\right)$ by $\left(ii\right)$, we get $\frac{N_{A}}{N_{B}}=e^{-\left(\lambda_{A}-\lambda_{B}\right)t}$ or $\frac{N_{A}}{N_{B}}=e^{-\left(5\lambda-\lambda\right)t}$ or $\left(\frac{1}{e}\right)^{2}=e^{-4\lambda t}$ or $\left(\frac{1}{e}\right)^{2}=\left(\frac{1}{e}\right)^{4\lambda t}$ $\Rightarrow 4\lambda t=2$ or $t=\frac{2}{4\lambda}=\frac{1}{2\lambda}$